How do you find the perimeter of a rectangle with the dimensions (x^2+2x-16) ft (length)
and (x+5) ft (width)

(2x²+6x-22) ft

Step-by-step explanation:

l = (x²+2x-16) ft

and w = (x+5) ft

the Perimeter =

2(x²+2x-16) +2(x+5)

= 2x²+4x-32+2x+10

= 2x²+6x-22

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.